Re: Can I get ComplexExpand to really work?

*To*: mathgroup at smc.vnet.net*Subject*: [mg14635] Re: Can I get ComplexExpand to really work?*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Wed, 4 Nov 1998 13:47:04 -0500*References*: <719f5p$lc6@smc.vnet.net> <71bkvu$pul@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Two observations: (1) on assumptions in Integrate(2) on your sequence. (1) The Integral As has already been pointed out by several people Integrate[ E^(I a b x^2),{x, -Infinity, Infinity}, Assumptions -> {a>0, b>0} ] If[Im[a*b] == 0, (Sqrt[Pi/2]*(1 + I*Sign[a*b]))/ (Sqrt[a]*Sqrt[b]), Integrate[E^(I*a*b*x^2), {x, -Infinity, Infinity}]] gives nearly what you want If we help Mathematica a bit we get Integrate[ E^(I a b x^2),{x, -Infinity, Infinity}, Assumptions -> {a b>0,a>0,b>0} ] ((1 + I)*Sqrt[Pi/2])/(Sqrt[a]*Sqrt[b]) Which, I guess, points up the need to internally extend the assumptions in the manner of the package Declare. (2) The sequence The sequence you gave was data = { {1, 0}, {2, 1/4}, {3, 1}, {4, 5/2}, {5, 5}, {6, 35/4}, {7, 14}, {8, 21}, {9, 30}, {10, 165/4}, {11, 55}}; and your answer, given later, was that this is given by the function f(N) = (N-1) N (N+1) / 24 Here is a Mathematica deduction of this Define p[n_] := InterpolatingPolynomial[Take[data, n], x] Look for where p[n] repeats fp = (n=1;FixedPoint[(++n; p[n]) &, 1]) (1/4 + (1/4 + 1/24*(-3 + x))*(-2 + x))*(-1 + x) Define f[x_] = Simplify[fp] 1/24*(-1 + x)*x*(1 + x) Check Last/@data == Table[f[n], {n, Length[data]}] True In a symbolic case, having used Mathematica to guess the formula, we might use induction to prove the result. --------------------- Allan Hayes Mathematica Training and Consulting www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 Topher Cawlfield wrote in message <71bkvu$pul at smc.vnet.net>... >Many thanks for the help! > >(I'm not quite sure how to best respond since this newsgroup seems to be >acting funny) > >Daniel Lichtblau, P.J. Hinton, and Lars Hohmuth suggested that I use the >"Assumptions" option with Integrate. Thanks, I didn't know about that! >This works well in the most simple case, but doesn't work quite right >in more general cases. > >Integrate[E^(I a b x^2), {x, -Infinity, Infinity}, Assumptions->{a>0 , >b>0}] > >Essentially returns Sqrt[Pi/(2 a b)] (1 + I Sign[a b]) > >This is progress, but I could make the argument that if a>0 and b>0, >then Sign[a b] = 1. > >If I use "Assumptions->{a b > 0}" instead, I get the more simple result. > >M. R. Schaferkotter suggested that I try the Declare.m package from >mathsource. I haven't tried this yet, but I will soon. > >--- >For the specific problem I was working on, I was finally able to use >Mathematica very successfully! I thought I would post it here since it >worked so well, and must have saved me at least a month of grinding >through algebra and making mistakes. Well, it didn't save me a month >really, it just made one technique practical whereas it wouldn't have >been otherwise. > >I want to use a Feynman path integral approach to find the transition >amplitude for an electron in a constant electric field. This boils >down to doing a lot of integrals, noticing a pattern, and deducing the >limiting case of performing an infinite number of integrals. > >I won't go into all the details, but I would end up needing to perform >integrals like: >Integrate[E^(I(a x^2 + b x y + c y^2 + d y z + e z^2 + f)), > {x, -Infinity, Infinity}, {y, -Infinity, Infinity}, {z, -Infinity, >Infinity}] >where a, b, c, d, e, and f are simple ratios of physical constants and >parameters in the problem. > >The best way to evaluate these by hand is to first "complete the square" >in x: >Integrate[E^(I(a (x+junk)^2 + more junk)),...] > >The value of "junk" doesn't matter since you can rescale x. The limits >of integration, being infinite, don't change. You can use the >integration rule I was getting at above. Then, repeat the process by >completing the square in y of "more junk", applying the integration >rule again, and moving on to z. > >I couldn't find any functions in mathematica that would complete the >square for me, but the CoefficientList function helped a lot. I made >my own function: > >CompleteTheSquare[p_, x_] := Module[{coeffs, fp}, > coeffs = CoefficientList[p, x]; > fp = coeffs[[3]] (x + coeffs[[2]] / (2 coeffs[[3]]) )^2; > parts = {coeffs[[3]], Simplify[p - fp], fp}; > fp + parts[[2]]] ] > >This function returns the modified polynomial, and also sets the symbol >'parts' to a list containing the various pieces of the result. > >I then defined a function that "did the integration": MyInt[a_] := >Sqrt[- Pi / a] >So that the integral of E^(a (x-b)^2 + c) is MyInt[a] E^c. > >And finally I made a function that did a Do loop over the integrals. >This was a bit of work, but the results were really quite nice. By >hand I could have only done 1 or 2 such integrals before tiring, and >that would have taken me hours. With this I could try any number. I >went up to 10, which just took a Mathematica a couple minutes. > >The morals of the story (IMO): (1) Use Mathematica for what it's good >at. (2) One still needs to be somewhat knowledgable of and practiced >with mathematical techniques. Even though Mathematica can do integrals >fairly well, it's still important to know how to do them "by hand". >This is a good thing. > >- Topher Cawlfield > >p.s. The fun part of the problem, besides the Mathematica programming, >was figuring out a strange number series that came up: > >What's f(N)? > >N | f(N) >---|---- >1 | 0 >2 | 1/4 >3 | 1 >4 | 5/2 >5 | 5 >6 | 35/4 >7 | 14 >8 | 21 >9 | 30 >10 | 165/4 >11 | 55 > >I'll post the answer tomorrow. These were coefficients in the answer. >To get one of them requires doing N-1 integrals symbolicly. >

**Follow-Ups**:**Re: Re: Can I get ComplexExpand to really work?***From:*Jurgen Tischer <jtischer@col2.telecom.com.co>

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